If a,b,c are in A.P then prove (a-c)²=4(b²-ac)
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Since a, b, c are in A.P so
b – a = c – b
b + b = c + a
2 b = c + a
Taking squares on both sides
(2 b) ² = (c + a)²
4b² = (c + a)²
Subtracting by 4ac on both sides, we get
4b² – 4ac = (c + a)² – 4 ac
4(b² – ac) = c² + a² + 2 ac – 4 ac
4(b² – ac) = c² + a² - 2 ac
4(b² – ac) = ( a – c )²
b – a = c – b
b + b = c + a
2 b = c + a
Taking squares on both sides
(2 b) ² = (c + a)²
4b² = (c + a)²
Subtracting by 4ac on both sides, we get
4b² – 4ac = (c + a)² – 4 ac
4(b² – ac) = c² + a² + 2 ac – 4 ac
4(b² – ac) = c² + a² - 2 ac
4(b² – ac) = ( a – c )²
shibushaw:
thank u so much
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